Question

Împărțind suma a 6 nr consecutive la 10,obtinem câtul 12 si restul 9.Sa se afle cele 6 nr.

Answer 1

a/10=12 rest 9
a(suma)=12*10 +9
a=129
suma=129
acum sa aflam numerele:
a+a+1+a+2+a+3+a+4+a+5=129
dam afara numerele 1,2,3,4,5.
129-1-2-3-4-5=114
114/6=19
a=19
a=20
a=21
a=22
a=23
a=24
 cele 6 numere sunt:19,20,21,22,23,24.
Verificare:19+20+21+22+23+24=129
129/10=12 rest 9

Answer 2

(6x+15)--suma numerelor
(6x+15):10=12rest9
6x+15=12x10+9
6x+15=129
6x=129-15
6x=114
x=114:6
x=19--primul nr
19+1=20--2nr
20+1=21--3nr
21+1=22-4nr
22+1=23--5nr
23+1=24--6nr